Calculus problem

ddxt2x31x2+t2dt=?\large \dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2+t^2}} \, dt = \, ?

Note by Ujjwal Mani Tripathi
3 years, 1 month ago

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Hint: Fundamental Theorem of Calculus.

Pi Han Goh - 3 years, 1 month ago

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yeah but i am still not able to get the required answer will you please post the solution to this question...

Ujjwal Mani Tripathi - 3 years ago

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The links pretty spells out the solution already. If you really can't solve it, why don't you post your attempt here and I'll see where you are stuck on.

Pi Han Goh - 3 years ago

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@Pi Han Goh i am getting the answer 3x2x2+x12 \dfrac{3x^2}{\sqrt{x^2 + x^12 }} it is x^12 .. in the denominator

Ujjwal Mani Tripathi - 3 years ago

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@Ujjwal Mani Tripathi Split the integral into 0x30t2 \int_0^{x^3} - \int_0^{t^2} . For the first integral, just apply the fundamental theorem of calculus. For the second theorem, you will get some constant number, so you can ignore it after you differentiate. Is this enough information for you?

Pi Han Goh - 3 years ago

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@Pi Han Goh Yeah that is enough

Ujjwal Mani Tripathi - 3 years ago

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@Ujjwal Mani Tripathi Is the answer correct?

Ujjwal Mani Tripathi - 3 years ago

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@Ujjwal Mani Tripathi you have to develop the steps...

Guillermo Templado - 3 years ago

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@Guillermo Templado do you mean that i really need to integrate the above expression first?

Ujjwal Mani Tripathi - 3 years ago

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@Ujjwal Mani Tripathi It's not necessary, but if you make the change t=x Sh ut = x \cdot \text{ Sh u} with xx constant, and use 1+Sh2(u)=Ch2(u)1 + Sh^2(u) = Ch^2(u) and dt=xCh u dudt = x \cdot \text{Ch u} \space du, the problemm looks like easy. Tomorrow, I promise you'll have my answer, now I'm very busy, Im' leaving my house. Tomorrow, I promise....

Guillermo Templado - 3 years ago

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@Guillermo Templado thanks man , appreciate the help

Ujjwal Mani Tripathi - 3 years ago

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@Ujjwal Mani Tripathi Ok, come on, please tell me if I'm wrong, I'm a person and due to this I make a lot of(too many) mistakes...

Proof 1 (attempt 1).-

ddxt2x31x2+t2dt=ddx(0x31x2+t2dt0t21x2+t2dt)=\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} \left( \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt - \int_{0}^{t^2} \dfrac 1{\sqrt{x^2 + t^2}} \, dt \right) = Let's call g(x)=x3g(x) = x^3 and f(x)=0x1x2+t2dtf(x) = \int_{0}^{x} \dfrac 1{\sqrt{x^2 + t^2}} \, dt Then, ddx0x31x2+t2dt=ddxf(g(x))=ddxf(x3)=3x2...\dfrac d{dx} \int_{0}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \dfrac d{dx} f(g(x)) = \dfrac d{dx} f(x^3) = 3x^2 \cdot ... Sorry, I'm getting a knot for myself...

Proof 2.-

ddxt2x31x2+t2dt= ()\dfrac d{dx} \int_{t^2}^{x^3} \dfrac 1{\sqrt{x^2 + t^2}} \, dt = \space (*) Let's make the change t=x Sh u t = x \cdot \text{ Sh u} , xx constant, dt=xCh u dudt = x \cdot \text{Ch u} \space du ()=ddxSh1(x Sh2u)Sh1x2± du=... (*) = \dfrac d{dx} \int_{Sh^{-1}(x \cdot \text{ Sh}^{2}u)}^{Sh^{-1} x^2} \, \pm \space du =... To be continued

Guillermo Templado - 3 years ago

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@Guillermo Templado Hmm it's getting better

Ujjwal Mani Tripathi - 3 years ago

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@Ujjwal Mani Tripathi I think the rest proof 2 should be simple, I'll finish it. I promise.... but you have to say me wheteher I'm right or not... It depends also if x>0x >0, x=0x = 0 or x<0x < 0 to concrete the ending. This question is giving me questions and suggestions... For example, does 0t dt\int_0^{t} \space dt make some sense?If the answer is yes, Is it tt?why?... I have another proof making the change u=x2+t2+tu = \sqrt{x^2 + t^2} + t... Even we can think first this question without the derivative ddx\frac{d}{dx}, I mean given xx possible concrete constants, x=1x =1, x=2x =2, x=100x = 100... and later, seeing the results, infering the last answer... And I'm also considering Pi's advice: Use fundamental theorem of Calculus... To be continued... Can you finish my 2nd proof, and I tell you what I think about your result? Come on, dare...

Guillermo Templado - 3 years ago

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@Ujjwal Mani Tripathi Make the change I have said to you... It must solve it quickly.. or use Fundamental theorem of calculus

Guillermo Templado - 3 years ago

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@Guillermo Templado can you provide the final answer , i am really stuck on this ?

Ujjwal Mani Tripathi - 3 years ago

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@Pi Han Goh @Guillermo Templado help will be welcomed .

Ujjwal Mani Tripathi - 3 years ago

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I agree with Pi Hang Goh, you can also make the change t=x\Sh(u)t = x\Sh(u) solve , the integral, and later derivate

Guillermo Templado - 3 years ago

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I mean t=xSh u t = x \cdot \text{Sh u}. I have some problems with my comments..

Guillermo Templado - 3 years ago

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Tomorrow, I rewiewed this, and if it's not solved. I'll try to solve it, I think I'm not going to have a lot problemms.. we'll see...

Guillermo Templado - 3 years ago

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